logarithm inequality Why $\frac{1}{n+1}<\log(n+1)-\log(n)<\frac{1}{n}$?
 A: $\frac{1}{n+1} < \log((n+1)/n)~ \Leftrightarrow ~ 1 < \log((1+1/n)^{n+1})$, which is equivalent to $e < (1+1/n)^{n+1}$. 
For the second inequality you get $e > (1+1/n)^n$. 
Both of these inequalities are true: The second one is true, because $e$ is defined as the limit of $(1+1/n)^n$ which is strictly monotonly increasing.
The first one is true because $(1+1/n)^{n+1}$ also converges to $e$ and is strictly monotonly decreasing.
A: Edmund Landau proves
$$1-\frac{1}{x} < \log x < x-1$$
Using the fact that
$$ \log x  =  \lim_{k\to \infty} k\left(x^{1/k}-1\right)$$  
He then states
$$\sum_{v=0}^{k-1} y^v {\leq \choose \geq} k \text{ ; for } {0 < x \leq 1 \choose x \geq 1}$$
Then put for $y>0$
$$y^k-1 = (y-1)\sum_{v=0}^{k-1} y^v \geq k(y-1)$$
Thus if $y = x^{\frac{1}{k}}$
$$x-1 \geq k\left(x^{1/k}-1\right) $$
For the second case 
$$\log{\frac{1}{x}} = - \log x$$
thus
$$ \log x = -\log{\frac{1}{x}} \geq 1-\frac{1}{x}$$
A: $\log(n+1) -\log(n) = \int_n^{n+1} {1 \over x}\,dx$. The integrand is between ${1 \over n+1}$ and ${1 \over n}$ and the interval of integration is of length $1$, so your integral will be between ${1 \over n + 1}*1 = {1 \over n + 1}$ and ${1 \over n}*1 = {1 \over n}$.
A: if n is integer>0 then you can do the following:
1a) $ \qquad  {1 \over n+1} < \log(1+n) - \log(n) < {1 \over n} $
rewrite:
1b) $ \qquad  {1 \over n+1} < \log(1+ {1 \over n} ) < {1 \over n} $      
First compare the two rhs terms of 1b). Exponentiation gives:
2) $ \qquad \qquad \qquad \qquad  1+ {1 \over n} < 1 + {1 \over n} + {1 \over n^2 *2!} + \ldots $ which is obvious for $n>0$    
Now compare the two lhs terms of 1b). Substitute $m$ for $n+1$ where now $m>1$:
3a) $ \qquad  {1 \over m} < \log (1 + {1 \over m-1}) $
exponentiate:
3b) $ \qquad \exp ({1 \over m}) <  (1 + {1 \over m-1}) = {m \over m-1} = {1 \over 1 - \frac1m} $
3c) $ \qquad  1 + {1 \over m} + {1 \over m^2 2! } + {1 \over m^3 3! }+\ldots <  1 + {1\over m} +{1\over m^2}+{1\over m^3}+\ldots  $  which, by termwise comparision, is obviously true , too.
